Can anyone present to me an elegant elementary proof of the relationship between the eigenvalues of a positive definite matrix and its Cholesky decomposition?

More formally, suppose $\mathbf{A}$ is an $n\times n$ positive definite matrix and let $\mathbf{A} = \mathbf{R}^\top \mathbf{R}$ be its Cholesky decomposition. Establish the relationship between the eigenvalues of $\mathbf{A}$ and that of $\mathbf{R}$.

EDIT (Additional remarks): My question specifically wants to find, if possible, an equation or function, say $f$, that relates the eigenvalues, i.e., $f\left(\lambda_i(\mathbf{R})\right) = \lambda_i(\mathbf{A})$, with uniqueness up to order being considered if necessary.


For a positive definite matrix $A$, with $Q$ as eigenvector matrix and $\Lambda$ as eigenvalue matrix, we have

$$ A = Q \Lambda Q^T $$

This can be rewritten as (since all eigenvalues of $A$ are positive) :

$$ A = (Q \sqrt{\Lambda}) (\sqrt{\Lambda} Q^T) $$

So for $A = R^TR$, $R$ can be a matrix such that,

$$ R = \sqrt{\Lambda} Q^T $$

Also we can multiply any orthogonal matrix $Q$ to this $R$ without changing the original $A = R^TR$ condition, because,

$$ A = (QR)^TQR = R^T(Q^TQ)R = R^TR $$

So rewriting $R$ as $Q\sqrt{\Lambda} Q^T$, we see that eigenvalues of $R$ are square roots of eigenvalues of $A$

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    $\begingroup$ How can you be sure that the $\mathbf{R}$ is an upper (or lower) triangular as should be the case of Cholesky? $\endgroup$ – venrey Sep 3 '18 at 16:43
  • $\begingroup$ Ah yes you're right. You can only write $R$ as $\sqrt {D} L^T$. So I think we can only comment on the sign of eigenvalues of $R$ matching those of $A$ but not the values themselves. $\endgroup$ – artha Sep 4 '18 at 4:14
  • $\begingroup$ Can you formalize your argument that, "the sign of eigenvalues of $\mathbf{R}$ matching those of $\mathbf{A}$ but not the values themselves"? $\endgroup$ – venrey Sep 4 '18 at 4:17

There is no such relation. If the spectrum of $A$ is a function of the spectrum of $R$, it would imply that $$ A=\pmatrix{1&0\\ t&1}\pmatrix{1&t\\ 0&1}=\pmatrix{1&t\\ t&t^2+1} $$ has a constant spectrum, but this is obviously not the case because our $A$ here has a non-constant trace.

In general, if $A=R^TR$ (regardless of whether this is a Cholesky decomposition or not) for a real square matrix $R$, the eigenvalues of $A$ are the squared singular values of $R$.


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